Determinant of a Matrix

Determinant of a Matrix

Determinant of a Matrix

Determinants originally appeared in the study of the linear equations. We shall, however, associate the notion of the determinant with matrices. For this purpose, we have to consider the square matrices of order 1, 2, 3 etc.

If A is a square matrix, its determinant is denoted by |A| or det.(A). And, the value of determinant is calculated differently for different ordered square matrices. Here is mentioned about the calculation of determinants of square matrices of order 1, 2 and 3.

 

Determinant of a 1×1 Matrix

Let A = (a11) be a square matrix of order 1×1. Then, its determinant |A| is given simply by the value of its single element a11.

i.e. |A| = |a11| = a11

For example: Let, A = (2), then

Determinant of A = |A| = |2| = 2

 

Determinant of a 2×2 Matrix

Let A = (■(a_11&a_12@a_21&a_22 )) be a square matrix of order 2×2. Then, its determinant |A| is given by the value of a11a22 – a12a21. i.e. |A| = |■(a_11&a_12@a_21&a_22 )| = a11a22 – a12a21		 For example: Let, A = (■(2&3@1&4)), then Determinant of A = |A| = |■(2&3@1&4)| = 2×4 - 3×1 = 8 – 3 = 5


Determinant of a 3×3 Matrix

Let A = (■(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )) be a square matrix of order 3×3. Then its determinant |A| is given by the value of a11A11 + a12A12 + a13A13. i.e. |A| = |■(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )| = a11A11 + a12A12 + a13A13 Where, A11, A12 and A13 are called the cofactors of the elements a11, a12 and a13 respectively and are given by the following calculations. A11 = (-1)1+1. |■(a_22&a_23@a_32&a_33 )| = a22a33 – a23a32 A12 = (-1)1+2. |■(a_21&a_23@a_31&a_33 )| = - (a21a33 – a23a31) A13 = (-1)1+3. |■(a_21&a_22@a_31&a_32 )| = a21a32 – a22a31 Here, the 2×2 determinant |■(a_22&a_23@a_32&a_33 )| is called minor of the element a11, which is obtained by omitting the 1st row and 1st column of 3×3 matrix A. Similarly, |■(a_21&a_23@a_31&a_33 )| is the minor of a12, and is obtained by omitting 1st row and 2nd column of A. And, |■(a_21&a_22@a_31&a_32 )| is the minor of a13, obtained by omitting 1st row and 3rd column of A.

Here is an easy pattern to find the minors of any element of a 3×3 matrix:

An easy pattern to find the minors of any element of a 3×3 matrix.

Note: Determinant of 3×3 matrix A can also be calculated by taking the products of the elements of any row (or column) and the corresponding cofactors. i.e.

|A| = a11A11 + a12A12 + a13A13

Or,

|A| = a21A21 + a22A22 + a23A23

Or,

|A| = a31A31 + a32A32 + a33A33

Or,

|A| = a11A11 + a21A21 + a31A31

Or,

|A| = a12A12 + a22A22 + a32A32

Or,

|A| = a13A13 + a23A23 + a33A33

All will give the same value for the determinant of A.

For example: Let, A = (■(1&2&0@2&3&1@5&4&2)) Now, A11 = (-1)1+1. |■(3&1@4&2)| = 6 – 4 = 2 A12 = (-1)1+2. |■(2&1@5&2)| = -(4 – 5) = 1		 A13 = (-1)1+3. |■(2&3@5&4)| = 8 – 15 = -7 A21 = (-1)2+1. |■(2&0@4&2)| = -(4 – 0) = - 4		 A22 = (-1)2+2. |■(1&0@5&2)| = 2 – 0 = 2		 A23 = (-1)2+3. |■(1&2@5&4)| = -(4 – 10) = 6 A31 = (-1)3+1. |■(2&0@3&1)| = 2 – 0 = 2		 A32 = (-1)3+2. |■(1&0@2&1)| = -(1 – 0) = - 1		 A33 = (-1)3+3. |■(1&2@2&3)| = 3 – 4 = - 1

Now,

|A| = a11A11 + a12A12 + a13A13 = 1×2 + 2×1 + 0×-7 = 4

|A| = a21A21 + a22A22 + a23A23 = 2×-4 + 3×2 + 1×6 = 4

|A| = a31A31 + a32A32 + a33A33 = 5×2 + 4×-1 + 2×-1 = 4

|A| = a11A11 + a21A21 + a31A31 = 1×2 + 2×-4 + 5×2 = 4

|A| = a12A12 + a22A22 + a32A32 = 2×1 + 3×2 + 4×-1 = 4

|A| = a13A13 + a23A23 + a33A33 = 0×-7 + 1×6 + 2×-1 = 4

Here, we got |A| = 4 in every calculations, so we can use any one of the above methods. Generally, we use the first method to calculate the determinant of a 3×3 matrix A. i.e.

|A| = a11A11 + a12A12 + a13A13


Mechanical Rule (Rule of Sarrus) 

A mechanical rule for finding the value of a third order determinant is as indicated below:-

Let |A| = |■(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )|  Add, to the three columns of a determinant, the first two columns respectively as the fourth and the fifth columns. |A| =  a11      a12      a13      a11      a12        +           a21      a22      a23      a21      a22           a31      a32      a33      a31      a32        –       = (a11a22a33 + a12a23a31 + a13a21a32) – (a31a22a13 + a32a23a11 + a33a21a12)

From the algebraic sum of the products of the elements of the diagonals pointing downwards, subtract the algebraic sum of the products of the elements of the diagonals pointing upwards.

This rule of expansion of a third order determinant is known as The Rule of Sarrus.

For example: Find the determinant of |A| = |■(1&2&-3@2&0&4@3&2&1)| by the rule of Sarrus. Solution: |A| = 1     2     -3     1     2 	              2     0       4     2     0 	              3     2       1     3     2                     = (0 + 24 – 12) – (0 + 8 + 4) 	            = 12 – 12 	            = 0

Note: This rule does not work for the determinants of the order greater than 3.

 

Properties of 3×3 Determinant

We have seen how to evaluate a 3×3 determinant. There are some properties of 3×3 determinant. Look at them and and their verification.

Prop. 1: The value of determinant is unaltered by interchanging its rows and columns.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )|  	  = a1|■(b_2&c_2@b_3&c_3 )| - b1|■(a_2&c_2@a_3&c_3 )| + c1|■(a_2&b_2@a_3&b_3 )|   = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)   = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 ……… (i) Now, |At|= |■(a_1&a_2&a_3@b_1&b_2&b_3@c_1&c_2&c_3 )|      = a1|■(b_2&b_3@c_2&c_3 )| - b1|■(a_2&a_3@c_2&c_3 )| + c1|■(a_2&a_3@b_2&b_3 )|      = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)      = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 …… (ii) From (i) and (ii), we get,|A| = |At| Hence verified.

Prop. 2: Interchanging any two adjacent rows (or columns) changes the sign of the determinant.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )|    = a1|■(b_2&c_2@b_3&c_3 )| - b1|■(a_2&c_2@a_3&c_3 )| + c1|■(a_2&b_2@a_3&b_3 )|   = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2)   = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 Let |B| be the determinant obtained by interchanging 1st and 2nd rows of |A|. ∴ |B|= |■(a_2&b_2&c_2@a_1&b_1&c_1@a_3&b_3&c_3 )|         = a2|■(b_1&c_1@b_3&c_3 )| – b2|■(a_1&c_1@a_3&c_3 )| + c2|■(a_1&b_1@a_3&b_3 )|        = a2(b1c3 – b3c1) – b2(a1c3 – a3c1) + c2(a1b3 – a3b1)        = a2b1c3 – a2b3c1 – a1b2c3 + a3b2c1 + a1b3c2 – a3b1c2        = - (a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1)        = - |A| Hence verified.

Prop. 3: If any two rows (or columns) of a determinant are identical, then the value of the determinant is zero.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_1&b_1&c_1@a_3&b_3&c_3 )| In this determinant first two rows are identical. Let |B| be the determinant obtained by interchanging first two rows in |A|. Then the determinant itself remains unchanged, but by prop. 2. its value is - |A| i.e. |B| = - |A| Hence, |A| = -|A| Or, |A| + |A| = 0 Or, 2|A| = 0 Or, |A| = 0 Hence verified.

Prop. 4: If all the elements of any row (or column) are multiplied by a constant k, then the value of determinant is multiplied by k.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )|             = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 (as obtained before) And, let |B| be the determinant obtained from A by multiplying the elements of the 1st column by a constant k (say). Then, |B| = |■(〖ka〗_1&b_1&c_1@〖ka〗_2&b_2&c_2@〖ka〗_3&b_3&c_3 )|      = ka1|■(b_2&c_2@b_3&c_3 )| - b1|■(〖ka〗_2&c_2@〖ka〗_3&c_3 )| + c1|■(〖ka〗_2&b_2@〖ka〗_3&b_3 )|      = ka1(b2c3 – b3c2) – b1(ka2c3 – ka3c2) + c1(ka2b3 – ka3b2)      = ka1b2c3 – ka1b3c2 – ka2b1c3 – ka3b1c2 + ka2b3c1 – ka3b2c1      = k(a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1)      = k|A| Hence verified.

Prop. 5: If each element of any row (or column) of a determinant is expressed as the sum of two terms, then the determinant can be expressed as the sum of determinants.

Verification:

Let |A| = |■(a_1+α&b_1&c_1@a_2+β&b_2&c_2@a_3+γ&b_3&c_3 )|, |B| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| and |C| = |■(α&b_1&c_1@β&b_2&c_2@γ&b_3&c_3 )|. Then we have to show that |A| = |B| + |C|. Here, |A| = |■(a_1+α&b_1&c_1@a_2+β&b_2&c_2@a_3+γ&b_3&c_3 )|      = (a1+α)|■(b_2&c_2@b_3&c_3 )| - (b1+β)|■(a_2&c_2@a_3&c_3 )| + (c1+γ)|■(a_2&b_2@a_3&b_3 )|      = a1|■(b_2&c_2@b_3&c_3 )| - b1|■(a_2&c_2@a_3&c_3 )| + c1|■(a_2&b_2@a_3&b_3 )| + α|■(b_2&c_2@b_3&c_3 )| - β|■(a_2&c_2@a_3&c_3 )| + γ|■(a_2&b_2@a_3&b_3 )|      = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| + |■(α&b_1&c_1@β&b_2&c_2@γ&b_3&c_3 )|      = |B| + |C| Hence verified.

Prop. 6: If to the elements of any row (or column) a multiple of any other row (or column) is added, the value of the determinant remains unaltered.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )|  And, let |B| be the determinant obtained by multiplying the elements of 2nd row of A by k and adding these products to the elements of the 1st row. Then, |B| = |■(a_1+ka_2&b_1+kb_2&c_1+kc_2@a_2&b_2&c_2@a_3&b_3&c_3 )|           (by prop. 5.)        = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| + |■(ka_2&kb_2&kc_2@a_2&b_2&c_2@a_3&b_3&c_3 )|            (by prop. 4.)                = |A| + k |■(a_2&b_2&c_2@a_2&b_2&c_2@a_3&b_3&c_3 )|               = |A| + k × 0 (∵ R1 = R2, by prop. 3.)       = |A| Hence verified.


Worked Out Examples

Example 1: Evaluate the determinant of the following matrices: 	A = (5) 	B = (■(1&2@3&4)) 	C = (■(3&-2&2@4&1&0@2&3&1)) Solution: Here, 	|A| = |5| = 5. 	|B| = |■(1&2@3&4)| = 1×4 - 2×3 = 4 – 6 = - 2. 	|C| = |■(3&-2&2@4&1&0@2&3&1)|         = 3|■(1&0@3&1)| – (-2)|■(4&0@2&1)| + 2|■(4&1@2&3)|        = 3(1 – 0) + 2(4 – 0) + 2(12 – 2)        = 3 × 1 + 2×4 + 2×10        = 3 + 8 + 20        = 31. Ans.
 

Example 2: If A = (■(1&-2@3&4)) and B = (■(-3&0@1&-2)), find the determinant of 5A – 2B + 3I. Solution: Here, A = (■(1&-2@3&4)), B = (■(-3&0@1&-2)) and I = (■(1&0@0&1)) Now,  5A – 2B + 3I  = 5(■(1&-2@3&4)) - 2(■(-3&0@1&-2)) + 3(■(1&0@0&1)) = (■(5&-10@15&20)) - (■(-6&0@2&-4)) + (■(3&0@0&3)) = (■(5+6+3&-10-0+0@15-2+0&20+4+3)) = (■(14&-10@13&27)) ∴ |5A – 2B + 3I| = |■(14&-10@13&27)| 	               = 14×27 + 10×13 	               = 378 + 130 	               = 508. Ans.

Example 3: If |■(x&2&3@-1&0&1@2&-2&0)| = 0, find the value of x. Solution: Here, |■(x&2&3@-1&0&1@2&-2&0)| = 0 Or,  x|■(0&1@-2&0)| - 2|■(-1&1@2&0)| + 3|■(-1&0@2&-2)| = 0 Or,  x(0 + 2) – 2(0 – 2) + 3(2 – 0) = 0 Or,  2x + 4 + 6 = 0 Or,  2x = -10 Or,  x = -5. Ans.


Example 4: Without expanding, show that |■(30&2&5@18&4&3@6&8&1)| = 0. Solution: Here, |■(30&2&5@18&4&3@6&8&1)| = |■(6×5&2&5@6×3&4&3@6×1&8&1)|  	   = 6 |■(5&2&5@3&4&3@1&8&1)|  	[Taking 6 common from Column 1] 	   = 6 × 0 [∵ Col. 1 = Col. 3] 	   = 0


Example 5: Find the value of |■(3&4&5@15&21&26@21&29&34)|. Solution: Here, Let, |A| = |■(3&4&5@15&21&26@21&29&34)| Multiplying 1st row by 5 and subtracting from 2nd row. And agan, multiplying  1st row by 7 and subtracting from 3rd row (i.e. Performing R2 – 5R1 and R3 – 7R1). We get |A| = |■(3&4&5@0&1&1@0&1&2)| = 3 |■(1&1@1&2)| - 0 |■(4&5@1&2)| + 0 |■(4&5@1&1)|        = 3(2 – 1) – 0(8 – 5) + 0(4 – 5)        = 3 × 1 – 0 + 0        = 3.


Example 6: Without expanding the determinant show that |■(b-c&b+c&b@c-a&c+a&c@a-b&a+b&a)| = 0. Solution: Here, LHS = |■(b-c&b+c&b@c-a&c+a&c@a-b&a+b&a)|         = |■(2b&b+c&b@2c&c+a&c@2a&a+b&a)|  	[Adding column 1 and 2 i.e. applying  	C1 + C2]         = 2 |■(b&b+c&b@c&c+a&c@a&a+b&a)|  	[Taking 2 common from C1]          = 2 × 0 [∵ C1 = C3]          = 0 = RHS.


Example 7: Show that |■(1&x&x^2@1&y&y^2@1&z&z^2 )| = (x – y)(y – z)(z – x). Solution: Here, LHS = |■(1&x&x^2@1&y&y^2@1&z&z^2 )|         = |■(1&x&x^2@0&y-x&y^2-x^2@0&z-x&z^2-x^2 )|  	[Applying R2 – R1, R3 – R1]         = |■(1&x&x^2@0×(y-x)&y-x&(y+x)(y-x)@0×(z-x)&z-x&(z+x)(z-x))|         = (y – x)(z – x) |■(1&x&x^2@0&1&y+x@0&1&z+x)|  	[Taking (y – x) common from R2 and  	(z – x) from R3]          = (y – x)(z – x) |■(1&y+x@1&z+x)|          = (y – x)(z – x)(z + x – y – x)          = (y – x)(z – x)(z – y)          = (x – y)(y – z)(z – x)          = RHS. Proved.

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